Three objects, A ∶ (a solid sphere), B ∶ (a thin circular disk) and C ∶ (a circular ring), each have the same mass M and radius R. They all spin with the same angular speed ω about their own symmetry axes. The amounts of work (W) required to bring them to rest, would satisfy the relation

Concept:

According to work energy theorem, work done by body is equal to change in its kinetic energy. A body in motion have two types of kinetic energy which are translational kinetic energy and rotational kinetic energy. Translation kinetic energy is due to its linear motion and rotational energy is due to circular motion of the body.

W = Change in Kinetic energy = K.E_T + K.E_R

where W is the work done

K.ET  is translation kinetic energy

K.ET = \(\frac{mv^2}{2}\), m and v is mass and speed of the body respectively.

K.ER is rotational kinetic energy

K.ER=\(\frac{Iw^2}{2}\), I and \(w\) is moment of inertia and angular speed of the body respectively.

Calculation:

Three objects, A ∶ (a solid sphere), B ∶ (a thin circular disk) and C ∶ (a circular ring), each have the same mass M and radius R.

 ω is angular speed

W is amount of work done

Work done required to bring them rest 

ΔW = ΔKE = K.ET  + K.ER

Since K.ET = 0

 \(Δ W = \frac{1}{2}I{ω ^2}\)

ΔW ∝ I for same ω

\({W_A}:{W_B}:{W_C} = \frac{2}{5}M{R^2}:\frac{1}{2}M{R^2}:M{R^2}\)

\( = \frac{2}{5}:\frac{1}{2}:1\)

\(= 4:5:10\)

⇒ W_C > W_B > W_A